Question: Simplify the following expression: $\dfrac{28y^5}{7y^2}$ You can assume $y \neq 0$.
Solution: $ \dfrac{28y^5}{7y^2} = \dfrac{28}{7} \cdot \dfrac{y^5}{y^2} $ To simplify $\frac{28}{7}$ , find the greatest common factor (GCD) of $28$ and $7$ $28 = 2 \cdot 2 \cdot 7$ $7 = 7$ $ \mbox{GCD}(28, 7) = 7 $ $ \dfrac{28}{7} \cdot \dfrac{y^5}{y^2} = \dfrac{7 \cdot 4}{7 \cdot 1} \cdot \dfrac{y^5}{y^2} $ $\phantom{ \dfrac{28}{7} \cdot \dfrac{5}{2}} = 4 \cdot \dfrac{y^5}{y^2} $ $ \dfrac{y^5}{y^2} = \dfrac{y \cdot y \cdot y \cdot y \cdot y}{y \cdot y} = y^3 $ $ 4 \cdot y^3 = 4y^3 $